 Capacitors & Dielectrics 02 OPENING QUESTIONS: A) What is the *simplist* definition for capacitance? B) Let's say we have a cylinder of length ℓ made of some sort of conducting material that is surrounded ("coaxial") by another cylinder, also of length ℓ . How would we go about finding the the electrical field as a function of r between the shells? (By the by, what does "coaxial" mean?) 1)Let's get someone up here to set this up 2) And someone else to continue 3) And someone to finish   ANSWER: In example 26.1, the book says "See formula 24.7"... which is hogwash, theres's NO WAY you are expected to remember every formula and every derivations... as it happens, the work ain't so hard, so let's do that now. Recall that we can use Gauss to find E field quite nicely: Φ = ∮E ∙ dA =∮Ecosθ dA = qin/εo Which we appropriately shorten to: ∮E ∙ dA = qin/εo The problem provides us the symetry we need so we quickly jump to: E∮∙ dA = qin/εo Did I mention symetry? Let's progress by subtituting λℓ in for q: EA = λℓ/εo Isolate for E: E= λℓ/Aεo We know the geometry (cylinder) so let's plug in the surface area of a cylinder (2πrℓ) for A: E= λℓ/2πrℓεo Cancel the like terms: E= λ/2πrεo Which gives us the value of E at any point r. YAYY! C) If we know the total charge Q on the enclosed cylinder, what OTHER additional information can we now find?     ANSWER: We recall that the definition of voltage is: ∆V=∫E∙ds (from A to B) We just found the value of E: E= λ/2πrεo So let's drop that in: ∆V=∫λ/2πrεo∙ds (from A to B) Pull out the constants and switch ds to dr since we've moving radially outward: ∆V=λ/(2πεo)∫dr/r (from A to B) dr/r is a well known value: -ln(r) ∆V=[-λ/(2πεo)]ln(r) evaluated from A to B but because you're all math wizards you recollect that ln(b) - ln(a) = ln(b/a) so: ∆V=[-λ/(2πεo)](ln(b/a) Recall ke = 1/4πεo so let's substitute that in: ∆V=(-2keλ)(ln(b/a) Now we're back to where the book was on page 781: We have ∆V and we have Q so we can find capacitance: C = Q/∆V Substituting in for ∆V from above (capacitance is always positive so we'll take the absolute value of our ∆V above: C = [Q/(2keλ)][ln(b/a)] We always get rid of λ as a last step, so let's recall λ = Q/ℓ: C = [Q/(2keQ/ℓ)][ln(b/a)] Cancel out the Q's, rearrange ℓ we get : C = ℓ/[(2ke)ln(b/a)] YIKES! I don't necessarily expect you to do the whole thing, but you should be able to follow along at the very least at first, and then reproduce it on your own... DO IT! OBJECTIVE:  I will be able to calculate the capacitance between two charged plate capacitors, cylindrical conducts and spherical conductors after today's class. WORDS/FORMULAE FOR TODAY TERMS: Capacitor - two charged surfaces that can store electrical energy CONSTANTS: ke = 8.987 x 109 Nm2/C2 ke = 1/4πεo UNITS: Capacitance = C (SI Units "farads" = f) C = Q/∆V farads are always positive capacitance measures ability of the system to store charge 1 farad is a MASSIVE amount of storage so we will typically talk in microfarads (μF = 10-6 F) or picofarads (pF = 10-12 F) FORMULAE: Capacitance: C = εoA/d (for a parallel plate capacitor) C = Q/∆V (generally) Electrical Field: For a parallel plate capacitor: E= σ/εo WORK O' THE DAY:  My annotated solution for capacitance of a spherical capicitor is below:   ════════════════════ Let's revisit a simple parallel plate capacitor: 1) Why is this 'device' called a capacitor? 2) How is it that the negatively charged plate becomes negatively charged? (think freely roaming electrons) 3) How is it that the positively charged plate becomes positively charged? (think freely roaming electrons) 4) If capacitance is a measure of how much electrical energy a capacitor can store, does that mean that farads are = to jules? Why or why not? Let's take a gander at THIS animation ════════════════════ Now let's take our FIRST gentle forray into circuits. An electrical circuit is called a circuit.... why????   An electrical circuit is ALWAYS composed of a power supply and wires. We aren't ever really concerned about a wire connected to the plus and minus parts of a battery because that's boring (there's nothing much for the electrons to do afterall). However, it starts to get interesting when we add one or more components to our circuit. We can add components to an electrical circuit in two ways, in series and in parallel. If we add only one component to the circuit, our math isn't all that interesting since we MUST add one end to the wire leading to the positive terminal and one end leading to the negative terminal in order for our circuit to be completed ... adding two or more components, then we care... Let's confine our conversations to capacitors (since that is our topic O' the day). If we drop a capacitor in a circuit with a battery we run a wire from the positive terminal lof the battery to one plate, and from the other plate to the negative terminal. That charges up the capacitor and we're in business. However, as noted before, adding two or more capacitors gets much more interesting. We can run them in series, or "end-to-end" as shown below: When we calculate the capacitance of the entire circuit, we add the reciprical of the capacitance of the capacitors in series. Let's say cap #1 above had a capacitance of 5.0 μf and cap #2 had a capacitance of 7.5 μf. To find the capacitance of the circuit we'd reverse 'em and add 'em: 1/5.0 μf + 1/7.5 μf = Don't forget to 1/x 'em when you're done We can also add the caps together in paralell: Which is a whole lot nicer when we're doing a circuit analysis since we just add the capicitance of both caps to get the total capacitance of the circuit. So if cap A is 5.0 μf and cap #2 had a capacitance of 7.5 μf then the total capacitance of the circuit is simply: 5.0 μf + 7.5 μf = 12.5μf   COURSEWORK: Problems: 26.4, 26.5, 26.6 (should be done!) 26.9, 26.13, 26.16, 26.20 ANSWERS:     STUDY GUIDE: